Exploring Number Theory: Permutations and Multiplicities

[noospace]

Hi all,

Consider the the number of distinct permutations of a collection of $N$ objects having multiplicities $n_1,\ldots,n_k$. Call this F.

Now arrange the same collection of objects into $k$ bins, sorted by type. Consider the set of permutations such that the contents of anyone bin after permutation are the same.

Can anyone help to convince me that the number of permutations which achieve this is also F? I believe that this is probably true but I'm unable to show it.

I've read elsewhere that $F = N!/(n_1!\cdots n_k!)$ which provides a starting point, but I'm not sure where to go from here.

 

[mighty2000]

Thank you for bringing up this interesting question. I can help you understand why the number of permutations that achieve the same contents in each bin is also equal to F.

First, let's define what F represents. F is the number of distinct permutations of a collection of N objects with multiplicities n_1, ..., n_k. This means that each unique arrangement of the objects in the collection counts as one permutation. So, if we have a collection of 3 objects with multiplicities 2, 1, 1, then the number of distinct permutations would be 3!/(2!1!1!) = 3.

Now, let's consider the set of permutations where the contents of each bin remain the same. This means that we are only rearranging the objects within each bin, but the overall contents of each bin remain unchanged. This is equivalent to arranging the objects in each bin separately, without considering the other bins.

Using the example above, if we have 2 bins with 2 objects in the first bin and 2 objects in the second bin, the number of permutations would be (2!/2!) * (2!/2!) = 1 * 1 = 1. This is the same as the number of distinct permutations, which is 3.

In general, for a collection of N objects with multiplicities n_1, ..., n_k, the number of permutations where the contents of each bin remain the same would be (n_1!/n_1!) * ... * (n_k!/n_k!) = 1 * ... * 1 = 1. This is also equal to F, as shown in the formula you mentioned (N!/(n_1!...n_k!)).

Therefore, we can conclude that the number of permutations that achieve the same contents in each bin is also equal to F. I hope this explanation helps to convince you of the truth of this statement. If you have any further questions, please don't hesitate to ask.